For otherwise, there would be a $k\in \mathbb{N}$ such that for every $\alpha \in X$ there is a $\beta > \alpha$ with $\lvert f(\beta) - f(\alpha)\rvert > 2^{-k}$. Then we could construct a sequence $(\gamma_n)$ with $\gamma_n < \gamma_{n+1}$ and $\lvert f(\gamma_{n+1}) - f(\gamma_n)\rvert > 2^{-k}$. But the sequence $(\gamma_n)$ converges to its supremum $\gamma \in X$, and hence